Unitary Matrix (original) (raw)
Last Updated : 13 Jun, 2026
A unitary matrix is a square matrix of complex numbers whose inverse is equal to its conjugate transpose. Equivalently, when a unitary matrix is multiplied by its conjugate transpose, the result is the identity matrix.
If U is a unitary matrix and Uᴴ (or U*) denotes its conjugate transpose, then:
- Uᴴ = U⁻¹
- UᴴU = UUᴴ = I
where , I is the identity matrix of the same order as U.
A unitary matrix is the complex-number equivalent of an orthogonal matrix. Its rows and columns are orthonormal, meaning they have unit length and are mutually perpendicular.
**Examples
- The matrix given below is a unitary matrix of order "2 × 2."
A_{2\times2} = \frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & 1\\ i & -i \end{array}\right]
- The matrix given below is a unitary matrix of order "3 × 3."
B_{3\times3} = \left[\begin{array}{ccc} i & 0 & 0\\ 0 & i & 0\\ 0 & 0 & i \end{array}\right]
Properties of a Unitary Matrix
The important properties of a unitary matrix are:
- A unitary matrix is always a square matrix.
- A unitary matrix is non-singular, which means its determinant is not zero.
- Every unitary matrix is invertible, and its inverse is equal to its conjugate transpose.
- The inverse of a unitary matrix is also unitary.
- The product of two unitary matrices of the same order is also a unitary matrix.
- A matrix is unitary if and only if its rows and columns are orthonormal.
- Every unitary matrix is diagonalizable.
- The absolute value of the determinant of a unitary matrix is always 1, i.e., |det(U)| = 1.
- The identity matrix is unitary.
- The transpose (or conjugate transpose) of a unitary matrix is also unitary.
- The modulus of every eigenvalue of a unitary matrix is 1, i.e., |λ| = 1.
**Note: The sum or difference of two unitary matrices is not necessarily a unitary matrix. Therefore, unitary matrices are closed under multiplication but not under addition or subtraction.
Solved Examples
**Example 1: Prove that the matrix given below is unitary.
U = \frac{1}{2}\left[\begin{array}{cc} 1+i & -1+i\\ 1+i & 1-i \end{array}\right]
**Solution:
To prove that the given matrix is unitary, we need to prove that UUH = I
Conjugate matrix of U =\frac{1}{2} \left[\begin{array}{cc} 1-i & -1-i\\ 1-i & 1+i \end{array}\right]
U^{H} = \frac{1}{2}\left[\begin{array}{cc} 1-i & 1-i\\ -1-i & 1+i \end{array}\right]
UU^{H} = \frac{1}{2}\left[\begin{array}{cc} 1+i & -1+i\\ 1+i & 1-i \end{array}\right]\times\frac{1}{2}\left[\begin{array}{cc} 1-i & 1-i\\ -1-i & 1+i \end{array}\right]
UU^{H} = \frac{1}{4} \left[\begin{array}{cc} [(1+i)(1-i)+(-1+i)(-1-i)] & [(1+i)(1-i)+(-1+i)(1+i)]\\{} [(1+i)(1-i)+(1-i)(-1-i)] & [(1+i)(1-i)+(i-i)(1+i)] \end{array}\right]
UU^{H} = \frac{1}{4}\left[\begin{array}{cc} [1+1+1+1] & [1+1-1-1]\\{} [1+1-1-1] & [1+1+1+1] \end{array}\right]
UU^{H} = \frac{1}{4}\left[\begin{array}{cc} 4 & 0\\ 0 & 4 \end{array}\right]
UU^{H} = \left[\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right] = I
Hence the given matrix is unitary.
**Example 2: Is the matrix given below a unitary matrix?
A = \left[\begin{array}{cc} i & 0\\ 0 & 1 \end{array}\right]
**Solution:
To prove that the given matrix is unitary, we need to prove that AAH = I
The conjugate matrix of U = \left[\begin{array}{cc} -i & 0\\ 0 & 1 \end{array}\right]
A^{H} = \left[\begin{array}{cc} -i & 0\\ 0 & 1 \end{array}\right]
AA^{H} = \left[\begin{array}{cc} i & 0\\ 0 & 1 \end{array}\right]\times\left[\begin{array}{cc} -i & 0\\ 0 & 1 \end{array}\right]
AA^{H} = \left[\begin{array}{cc} [(i\times-i)+0] & [0+0]\\{} [0+0] & [0+1] \end{array}\right]
AA^{H} = \left[\begin{array}{cc} -i^{2} & 0\\ 0 & 1 \end{array}\right]
AA^{H} = \left[\begin{array}{cc} -(-1) & 0\\ 0 & 1 \end{array}\right] = \left[\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right] = I
Hence the given matrix is unitary.
**Example 3: Prove that the absolute value of the determinant of a unitary matrix is one.
P =\frac{1}{3} \left[\begin{array}{cc} 2-i & 2\\ 2 & 2+i \end{array}\right]
**Solution:
P = \left[\begin{array}{cc} \frac{2-i}{3} & \frac{-2}{3}\\ \frac{2}{3} & \frac{2+i}{3} \end{array}\right]
|P| = \left|\begin{array}{cc} \frac{2-i}{3} & \frac{-2}{3}\\ \frac{2}{3} & \frac{2+i}{3} \end{array}\right|
|P| = (\frac{2-i)}{3})(\frac{2+i}{3})-(\frac{-2}{3})(\frac{2}{3})
|P| = (\frac{4-i^{2}}{9})+(\frac{4}{9})
|P| = (\frac{4-(-1)}{9})+(\frac{4}{9})
|P| = \frac{5}{9}+\frac{4}{9}= \frac{(5+4)}{9}= \frac{9}{9}= 1
Hence proved.